Nelson Functions 11
Nelson Functions 11
1st Edition
Chris Kirkpatrick, Marian Small
ISBN: 9780176332037
Textbook solutions

All Solutions

Page 534: Practice Questions

Exercise 1
Step 1
1 of 7
[begin{gathered}
{text{For simple interest, the formula is}} hfill \
hfill \
A = P + I = Pleft( {1 + rt} right) hfill \
hfill \
{text{where}} hfill \
A; = {text{ amount of investment after }}t{text{ years}} hfill \
P = {text{principal amount invested}} hfill \
r = {text{ annual interest rate}} hfill \
t{text{ = number of years}} hfill \
I = {text{ }}Prt = {text{interest after }}t{text{ years }} hfill \
end{gathered} ]
Step 2
2 of 7
[begin{gathered}
{text{a}}{text{.)}},, hfill \
P = 3500 hfill \
r = 0.06 hfill \
t = 10{text{ years}} hfill \
I = Prt = left( {3500} right)left( {0.06} right)left( {10} right) = $ 2100 hfill \
A = P + I = 3500 + 2100 = $ 5600 hfill \
end{gathered} ]
Step 3
3 of 7
[begin{gathered}
{text{b}}{text{.) }} hfill \
P = 15000 hfill \
r = 0.11 hfill \
t = 3{text{ years}} hfill \
I = Prt = left( {15000} right)left( {0.11} right)left( 3 right) = $ 495 hfill \
A = P + I = 15000 + 495 = $ 15495 hfill \
end{gathered} ]
Step 4
4 of 7
[begin{gathered}
{text{c}}{text{.)}} hfill \
P = 280 hfill \
r = 0.032 hfill \
t = 34{text{ months }} = frac{{34}}{{12}}{text{ years}} hfill \
I = left( {280} right)left( {0.032} right)left( {frac{{34}}{{12}}} right) = $ 22.67 hfill \
A = P + I = 280 + 22.67 = $ 302.67 hfill \
end{gathered} ]
Step 5
5 of 7
[begin{gathered}
{text{d}}{text{.)}} hfill \
P = 850 hfill \
r = 0.29 hfill \
t = 100{text{ weeks = }}frac{{100}}{{52}}{text{ years}} hfill \
I = 850left( {0.29} right)left( {frac{{100}}{{52}}} right) = $ 474.04 hfill \
A = 850 + 474.04 = $ 1324.04 hfill \
end{gathered} ]
Step 6
6 of 7
[begin{gathered}
{text{e}}{text{.) }} hfill \
P = 21000 hfill \
r = 0.073 hfill \
t = 42{text{ days }} = frac{{42}}{{365}}{text{ years}} hfill \
I = 21000left( {0.073} right)left( {frac{{42}}{{365}}} right) = $ 176.4 hfill \
A = 21000 + 176.4 = $ 21176.4 hfill \
end{gathered} ]
Result
7 of 7
begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
& Interest & Total Amount \ hline
a) & $2100 & $5600 \ hline
b) & $495 & $15495 \ hline
c) & $22.67 & $302.67 \ hline
d) & $474.04 & $1324.04 \ hline
e) & $176.40 & $21176.40 \ hline
end{tabular}
end{table}
Exercise 2
Step 1
1 of 5
[begin{gathered}
{text{For simple interest, the formula is}} hfill \
hfill \
A = P + I = Pleft( {1 + rt} right) hfill \
hfill \
{text{where}} hfill \
A; = {text{ amount of investment after }}t{text{ years}} hfill \
P = {text{principal amount invested}} hfill \
r = {text{ annual interest rate}} hfill \
t{text{ = number of years}} hfill \
I = {text{ }}Prt = {text{interest after }}t{text{ years }} hfill \
end{gathered} ]
Step 2
2 of 5
a.) Calculate the interest $I$

$I=Prt$

$r=dfrac{I}{Pt}=dfrac{11.25}{2500(1/12)}=0.054=5.4%$

Step 3
3 of 5
b.) Calculate the total amount after 7 years

$A=P(1+rt)=2500(1+0.054times 7)=$3445$

Step 4
4 of 5
c.) Find $t$ such that $A=2P$

$2P=P(1+rt)$

$2=1+rt$

$t=dfrac{1}{r}=dfrac{1}{0.054}=18.52$ years

Result
5 of 5
a) $5.4%$

b) $$3445$

c) $18.52$ years

Exercise 3
Step 1
1 of 5
[begin{gathered}
{text{For simple interest, the formula is}} hfill \
hfill \
A = P + I = Pleft( {1 + rt} right) hfill \
hfill \
{text{where}} hfill \
A; = {text{ amount of investment/debt after }}t{text{ years}} hfill \
P = {text{principal amount invested/loaned}} hfill \
r = {text{ annual interest rate}} hfill \
t{text{ = number of years}} hfill \
I = {text{ }}Prt = {text{interest after }}t{text{ years }} hfill \
end{gathered} ]
Step 2
2 of 5
a.) The amount that Karl borrowed is the point in the graph that corresponds to the initial year (1st year) which is $$5000$
Step 3
3 of 5
b.) The annual interest can be calculated from any two points in the graph.

If we consider

$A_1=5000$

$A_3=6000$

Then the interest rate is

$I=Prtimplies 6000-5000=(5000)(r)(2)$

$$
r=0.1=10%
$$

Step 4
4 of 5
c.) Find $t$ such that $A=20,000$

$A=P(1+rt)$

$1+rt=dfrac{A}{P}$

$t=dfrac{1}{r}left(dfrac{A}{P}-1right)$

$t=dfrac{1}{0.1}left(dfrac{20000}{5000}-1right)=30$ years

Result
5 of 5
a) $$5000$

b) $10%$

c) 30 years

Exercise 4
Step 1
1 of 5
[begin{gathered}
{text{The value of an investment in compound interest is}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
hfill \
{text{where }} hfill \
hfill \
P{text{ = principal amount}} hfill \
i = {text{ interest rate per compounding period}} hfill \
{text{m }} = {text{ number of compounding periods}} hfill \
hfill \
{text{If the annual interest rate }}r{text{ and number of years }}t{text{ is given}} hfill \
hfill \
i = frac{r}{n} hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 5
[begin{gathered}
{text{In this case, }} hfill \
P = 4350 hfill \
r = 7.6% /a hfill \
{text{compounded quarterly}} hfill \
hfill \
{text{We shall find how long it takes for }}A = 10000 hfill \
A = P{left( {1 + i} right)^m} hfill \
10000 = 4350{left( {1 + frac{{0.076}}{4}} right)^{4t}} hfill \
{left( {1.019} right)^{4t}} = frac{{200}}{{87}} hfill \
end{gathered} ]
Step 3
3 of 5
Exercise scan
Step 4
4 of 5
Therefore, it would take around $boxed{bf{11;years}}$
Result
5 of 5
11 years
Exercise 5
Step 1
1 of 6
[begin{gathered}
{text{The value of an investment in compound interest is}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
hfill \
{text{where }} hfill \
hfill \
P{text{ = principal amount}} hfill \
i = {text{ interest rate per compounding period}} hfill \
{text{m }} = {text{ number of compounding periods}} hfill \
hfill \
{text{If the annual interest rate }}r{text{ and number of years }}t{text{ is given}} hfill \
hfill \
i = frac{r}{n} hfill \
hfill \
m = n times t hfill \
hfill \
{text{The interest charged is}} hfill \
hfill \
I = A – P hfill \
end{gathered} ]
Step 2
2 of 6
[begin{gathered}
{text{a}}{text{.) In this case, }} hfill \
P = 4300 hfill \
r = 9.1% /a hfill \
{text{compounded annually}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
A = 4300{left( {1 + 0.091} right)^8} = $ 8631.11 hfill \
hfill \
I = A – P hfill \
I = 8631.11 – 4300 = $ 4331.11; hfill \
end{gathered} ]
Step 3
3 of 6
[begin{gathered}
{text{b}}{text{.) In this case, }} hfill \
P = 500 hfill \
r = 10.4% /a hfill \
{text{compounded semi – annually}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
A = 500{left( {1 + frac{{0.104}}{2}} right)^{2left( {11.5} right)}} = $ 1604.47 hfill \
hfill \
I = A – P hfill \
I = 1604.47 – 500 = $ 1104.47; hfill \
end{gathered} ]
Step 4
4 of 6
[begin{gathered}
{text{c}}{text{.) In this case, }} hfill \
P = 25000 hfill \
r = 6.4% /a hfill \
{text{compounded quarterly}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
A = 25000{left( {1 + frac{{0.064}}{4}} right)^{4left( 3 right)}} = $ 30245.76 hfill \
hfill \
I = A – P hfill \
I = 30245.76 – 25000 = $ 5245.76 hfill \
end{gathered} ]
Step 5
5 of 6
[begin{gathered}
{text{d}}{text{.) In this case, }} hfill \
P = 307 hfill \
r = 27.6% /a hfill \
{text{compounded monthly}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
A = 307{left( {1 + frac{{0.276}}{{12}}} right)^{12left( {2.5} right)}} = $ 607.31 hfill \
hfill \
I = A – P hfill \
I = 607.31 – 307 = 300.31 hfill \
end{gathered} ]
Result
6 of 6
begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
& Total Amount & Interest \ hline
a) & $8631.11 & $4331.11 \ hline
b) & 1604.47 & $1104.47 \ hline
c) & $30245.76 & $5245.76 \ hline
d) & $607.31 & $300.31 \ hline
end{tabular}
end{table}
Exercise 6
Step 1
1 of 4
[begin{gathered}
{text{The value of an investment in compound interest is}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
hfill \
{text{where }} hfill \
hfill \
P{text{ = principal amount}} hfill \
i = {text{ interest rate per compounding period}} hfill \
{text{m }} = {text{ number of compounding periods}} hfill \
hfill \
{text{If the annual interest rate }}r{text{ and number of years }}t{text{ is given}} hfill \
hfill \
i = frac{r}{n} hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 4
[begin{gathered}
{text{a.) In this case,}} hfill \
{text{interest after 1st year = $ 400}} hfill \
{text{interest after 2nd year = $ 432}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
400 = P{left( {1 + i} right)^1} hfill \
432 = P{left( {1 + i} right)^2} hfill \
{text{Equate }}P{text{ from both equations}} hfill \
hfill \
frac{{400}}{{left( {1 + i} right)}} = frac{{432}}{{{{left( {1 + i} right)}^2}}} hfill \
400 = frac{{432}}{{1 + i}} hfill \
1 + i = frac{{432}}{{400}} hfill \
i = frac{{432}}{{400}} – 1 hfill \
i = 0.08 = boxed{{mathbf{8.00% /a}}} hfill \
end{gathered} ]
Step 3
3 of 4
[begin{gathered}
{text{b}}{text{.) The amount that she invested is obtained as}} hfill \
A = P{left( {1 + i} right)^m} hfill \
P + 400 = Pleft( {1 + 0.08} right) hfill \
P = frac{{400}}{{0.08}} = boxed{{mathbf{$ 5000}}.{mathbf{00}}} hfill \
end{gathered} ]
Result
4 of 4
a) $8.00%/a$

b) $$5000$

Exercise 7
Step 1
1 of 3
[begin{gathered}
{text{The value of an investment in compound interest is}} hfill \
hfill \
A = P{left( {1 + i} right)^m} hfill \
hfill \
{text{where }} hfill \
hfill \
P{text{ = principal amount}} hfill \
i = {text{ interest rate per compounding period}} hfill \
{text{m }} = {text{ number of compounding periods}} hfill \
hfill \
{text{If the annual interest rate }}r{text{ and number of years }}t{text{ is given}} hfill \
hfill \
i = frac{r}{n} hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{In this case,}} hfill \
{text{Total cost}} = 2942.37 hfill \
{text{downpayment = }}850 hfill \
t{text{ = 18 months}} hfill \
hfill \
{text{The amount that Vlad financed is}} hfill \
{text{the cost minus the downpayment}} hfill \
P = 2942.37 – 850 = 2092.37 hfill \
hfill \
{text{At the end of 18 months,}} hfill \
A = 2147.48 hfill \
hfill \
{text{We shall determine }}i hfill \
A = P{left( {1 + i} right)^m} hfill \
2147 = 2092.37{left( {1 + i} right)^{18}} hfill \
i = {left( {frac{{2147.48}}{{2092.37}}} right)^{1/18}} – 1 hfill \
i = 0.001445 hfill \
hfill \
{text{Since the compounding was monthly}} hfill \
r = 12 times 0.001445 = 0.0173 hfill \
boxed{{mathbf{r = 1}}{mathbf{.73% /a}}} hfill \
end{gathered} ]
Result
3 of 3
$$
1.73%/a
$$
Exercise 8
Step 1
1 of 3
[begin{gathered}
{text{The present value of an investment/loan is}} hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
FV = {text{future value}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding period}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi-annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{a}}{text{.) }} hfill \
r = 6.7% = 0.067 hfill \
{text{compounded annually}} hfill \
t = {text{ 5 years}} hfill \
FV = 8000 hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} = frac{{8000}}{{{{left( {1 + frac{{0.067}}{1}} right)}^{5left( 1 right)}}}} = boxed{{mathbf{$ 5784}}{mathbf{.53}}} hfill \
hfill \
{text{b}}{text{.) }} hfill \
r = 8.8% = 0.088 hfill \
{text{compounded semi – annually}} hfill \
t = 2.5{text{ years}} hfill \
FV = 1,280 hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} = frac{{1280}}{{{{left( {1 + frac{{0.088}}{2}} right)}^{2left( {2.5} right)}}}} = boxed{{mathbf{$ 1032}}{mathbf{.07}}} hfill \
hfill \
{text{c}}{text{.)}} hfill \
r = 5.6% = 0.056 hfill \
{text{compounded quarterly}} hfill \
t = 8{text{ years}} hfill \
FV = 100,000 hfill \
hfill \
PV = frac{{100,000}}{{{{left( {1 + frac{{0.056}}{4}} right)}^{4left( 8 right)}}}} = boxed{{mathbf{$ 64089}}{mathbf{.29}}} hfill \
hfill \
{text{d}}{text{.)}} hfill \
r = 24.6% = 0.246 hfill \
{text{compounded monthly}} hfill \
t = 1.5{text{ years}} hfill \
FV = {text{ $ 850}} hfill \
hfill \
PV = frac{{850}}{{{{left( {1 + frac{{0.246}}{{12}}} right)}^{12left( 1.5 right)}}}} = boxed{{mathbf{$ 589}}{mathbf{.91}}} hfill \
end{gathered} ]
Result
3 of 3
a) $5784.53

b) $1032.07

c) $64089.29

d) $589.91

Exercise 9
Step 1
1 of 3
[begin{gathered}
{text{The present value of an investment/loan is}} hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
FV = {text{future value}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding period}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi-annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{In this case,}} hfill \
hfill \
FV = 847.53 hfill \
r = 9.6% /a hfill \
{text{compounded monthly}} hfill \
t = 2.5{text{ years}} hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} = frac{{847.53}}{{{{left( {1 + frac{{0.096}}{{12}}} right)}^{12left( {2.5} right)}}}} = boxed{{mathbf{$ 667}}{mathbf{.33}}} hfill \
hfill \
{text{Therefore, Roberto borrowed $667}}{text{.33}} hfill \
end{gathered} ]
Result
3 of 3
$$
$667.33
$$
Exercise 10
Step 1
1 of 3
[begin{gathered}
{text{The present value of an investment/loan is}} hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
FV = {text{future value}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding period}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi-annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{In this case,}} hfill \
hfill \
FV = 2262.70 hfill \
PV = 1650 hfill \
{text{compounded annually}} hfill \
t = 3{text{ years}} hfill \
hfill \
PV = frac{{FV}}{{{{left( {1 + i} right)}^m}}} hfill \
1650 = frac{{2262.70}}{{{{left( {1 + i} right)}^3}}} hfill \
{left( {1 + i} right)^3} = frac{{2262.70}}{{1650}} hfill \
i = {left( {frac{{2262.70}}{{1650}}} right)^{1/3}} – 1 hfill \
i = 0.111 = boxed{{mathbf{11}}{mathbf{.1% /a}}} hfill \
hfill \
{text{Therefore, an annual interest rate of 11}}{text{.1% would yield}} hfill \
{text{the same results}}{text{.}} hfill \
end{gathered} ]
Result
3 of 3
$$
11.1%/a
$$
Exercise 11
Step 1
1 of 5
[begin{gathered}
{text{The future value of an annuity is}} hfill \
hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
FV = {text{ future value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 5
[begin{gathered}
{mathbf{a)}}{text{ }} hfill \
R = 2500 hfill \
r = 7.6% = 0.076 hfill \
{text{compounded annually}} hfill \
t = 12{text{ years}} hfill \
i = 0.076 hfill \
m = {text{ 12}} times {text{1}} = 12 hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} = 2500 cdot frac{{{{left( {1 + 0.076} right)}^{12}} – 1}}{{0.076}} hfill \
FV = $ 46332.35 hfill \
hfill \
I = ,FV – m cdot R hfill \
I = 46332.35 – 12left( {2500} right) = $ 16332.35 hfill \
end{gathered} ]
Step 3
3 of 5
[begin{gathered}
{mathbf{b}}.) hfill \
R = 500 hfill \
r = 7.2% = 0.072 hfill \
{text{semi – annually}} hfill \
t = 9.5{text{ years}} hfill \
i = 0.072/2 = 0.036 hfill \
m = {text{ 9}}{text{.5}} times {text{2}} = 19 hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} = 500 cdot frac{{{{left( {1 + 0.036} right)}^{19}} – 1}}{{0.036}} hfill \
FV = $ 13,306.97 hfill \
hfill \
I = FV – m cdot R hfill \
I = 13,306.97 – 19left( {500} right) = 3,806.97 hfill \
end{gathered} ]
Step 4
4 of 5
[begin{gathered}
{mathbf{c}}.) hfill \
R = 2500 hfill \
r = 4.3% = 0.043 hfill \
{text{quarterly}} hfill \
t = 3{text{ years}} hfill \
i = 0.043/4 = 0.01075 hfill \
m = {text{ 4}} times {text{3}} = 12 hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} = 2500 cdot frac{{{{left( {1 + 0.01075} right)}^{12}} – 1}}{{0.01075}} hfill \
FV = $ 31,838.87 hfill \
hfill \
I = FV – m cdot R hfill \
I = 31,838.87 – 12left( {2500} right) = $ 1838.87 hfill \
end{gathered} ]
Result
5 of 5
begin{table}[]
defarraystretch{1.6}%
begin{tabular}{|l|l|l|}
hline
& Future Value & Interest \ hline
a) & $46,332.35 & $16,332.35 \ hline
b) & $13,306.97 & $3806.97 \ hline
c) & $31,838.87 & $1838.87 \ hline
end{tabular}
end{table}
Exercise 12
Step 1
1 of 5
[begin{gathered}
{text{The future value of an annuity is}} hfill \
hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
FV = {text{ future value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 5
[begin{gathered}
{text{In this case,}} hfill \
R = 1500 hfill \
{text{quarter payments}} hfill \
r = 4.4% /a hfill \
hfill \
{text{We shall}};{text{find how long it takes to reach}} hfill \
FV = 100,000 hfill \
hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
{text{100,000 = 1500}} cdot frac{{{{left( {1 + frac{{0.044}}{4}} right)}^{4t}} – 1}}{{left( {0.044/4} right)}} hfill \
{left( {1.011} right)^{4t}} = frac{{26}}{{15}} hfill \
end{gathered} ]
Step 3
3 of 5
Exercise scan
Step 4
4 of 5
Thus, the goal will be achieved in 12.57 years or 12 years and 7 months.
Result
5 of 5
12 years and 7 months
Exercise 13
Step 1
1 of 3
[begin{gathered}
{text{The future value of an annuity is}} hfill \
hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
FV = {text{ future value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
hfill \
m = n times t hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{In this case,}} hfill \
r = 9% /a hfill \
{text{compounded monthly}} hfill \
t = 6{text{ years}} hfill \
FV = 25,000 hfill \
i = 0.09/12 = 0.0075 hfill \
m = 6 times 12 = 72 hfill \
hfill \
{text{We need to find }}R hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
R = frac{{FV cdot i}}{{{{left( {1 + i} right)}^m} – 1}} hfill \
R = frac{{25,000left( {0.0075} right)}}{{{{left( {1 + 0.0075} right)}^{72}} – 1}} = boxed{{mathbf{$ 263}}{mathbf{.14}}} hfill \
{text{Ernie has to deposit $ 263}}{text{.14 every month}}{text{.}} hfill \
end{gathered} ]
Result
3 of 3
$$
$263.14
$$
Exercise 14
Step 1
1 of 3
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{We shall calculate the amount of loan and interest}}{text{.}} hfill \
{mathbf{a)}} hfill \
PV = R cdot frac{{1 – {{left( {1 + frac{r}{n}} right)}^{nt}}}}{{r/n}} = 450 cdot frac{{1 – {{left( {1 + frac{{0.051}}{1}} right)}^{ – 6left( 1 right)}}}}{{0.051}} = $ 2276.78 hfill \
I = R cdot m – PV = left( {450} right)left( {12} right) – 2276.78 = $ 423.22 hfill \
{mathbf{b}}) hfill \
PV = 2375 cdot frac{{1 – {{left( {1 + frac{{0.092}}{2}} right)}^{ – 4.5left( 2 right)}}}}{{0.092/2}} = $ 17,185.88 hfill \
I = R cdot m – PV = left( {2375} right)left( {4.5 times 2} right) – 17,185.88 = $ 4189.12 hfill \
{mathbf{c}}) hfill \
PV = 185.73 cdot frac{{1 – {{left( {1 + frac{{0.128}}{4}} right)}^{ – 4left( {3.5} right)}}}}{{0.128/4}} = $ 2069.71 hfill \
I = R cdot m – PV = left( {185.73} right)left( {3.5 times 4} right) – 2069.701 = $ 530.52 hfill \
hfill \
{mathbf{d}}) hfill \
PV = 105.27 cdot frac{{1 – {{left( {1 + frac{{0.192}}{{12}}} right)}^{ – 12left( {1.5} right)}}}}{{0.192/12}} = $ 1635.15 hfill \
I = R cdot m – PV = left( {105.27} right)left( {12 times 1.5} right) – 1635.15 = $ 259.71 hfill \
end{gathered} ]
Result
3 of 3
begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
& Amount of Loan & Interest \ hline
a) & $2276.78 & $423.22 \ hline
b) & $17,185.88 & $4189.12 \ hline
c) & $2069.71 & $530.52 \ hline
d) & $1635.15 & $259.71 \ hline
end{tabular}
end{table}
Exercise 15
Step 1
1 of 3
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{mathbf{a)}}{text{ In this case,}} hfill \
PV = 136,000 hfill \
{text{monthly payments}} hfill \
t = 20{text{ years}} hfill \
r = 6.6% /a hfill \
i = frac{{0.066}}{{12}} = 0.0055 hfill \
m = 20 times 12 = 240 hfill \
hfill \
{text{We shall calculate Paul’s monthly payment}} hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
R = frac{{PV cdot i}}{{1 – {{left( {1 + i} right)}^{ – m}}}} = frac{{136,000left( {0.0055} right)}}{{1 – {{left( {1 + 0.0055} right)}^{ – 240}}}} hfill \
boxed{{mathbf{R = $ 1022}}{mathbf{.00}}} hfill \
hfill \
{mathbf{b)}}{text{ The interest charged is}} hfill \
I = R cdot m – PV hfill \
I = left( {1022} right)left( {240} right) – 136,000 hfill \
boxed{{mathbf{I = $ 109,280}}} hfill \
end{gathered} ]
Result
3 of 3
a) $R=$1022.00$

b) $I=$109,280$

Exercise 16
Step 1
1 of 5
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 5
[begin{gathered}
{text{ In this case,}} hfill \
PV = $ 611.03 hfill \
R = $ 26.17 hfill \
{text{monthly payments}} hfill \
t = 2.5{text{ years}} hfill \
hfill \
{text{We shall calculate the annual interest rate}} hfill \
i = frac{r}{{12}} hfill \
m = 12 times 2.5 = 30 hfill \
hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i},, Rightarrow frac{{PV}}{R} = frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
frac{{611.03}}{{26.17}} = frac{{1 – {{left( {1 + i} right)}^{ – 30}}}}{i} hfill \
end{gathered} ]
Step 3
3 of 5
Exercise scan
Step 4
4 of 5
Since the monthly interest is $0.017$, the annual interest rate is

$$
r=0.017times 12=20.40%
$$

Result
5 of 5
$$
r=20.40%
$$
Exercise 17
Step 1
1 of 3
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
PV = {text{ present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi-annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{In this case,}} hfill \
{text{total cost}} = $ 1875.47 hfill \
{text{downpayment = }},{text{$ 650}} hfill \
r = 6.6% /a hfill \
{text{compounded monthly}} hfill \
t = 4{text{ years}} hfill \
hfill \
{text{We shall find the monthly payments}}{text{.}} hfill \
hfill \
{text{The amount to be financed is}},{text{the total cost}} hfill \
{text{minus the downpayment}} hfill \
PV = 1875.47 – 650 = 1225.47 hfill \
hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i},, Rightarrow R = frac{{PV cdot i}}{{1 – {{left( {1 + i} right)}^{ – m}}}} hfill \
hfill \
R = frac{{1225.47left( {frac{{0.066}}{{12}}} right)}}{{1 – {{left( {1 + frac{{0.066}}{{12}}} right)}^{ – 4left( {12} right)}}}} = boxed{{mathbf{$ 29}}{mathbf{.12}}} hfill \
hfill \
{text{Therefore, Chantal has to pay $ 29}}{text{.12 per month}}{text{.}} hfill \
end{gathered} ]
Result
3 of 3
$$
r=20.40%
$$
Exercise 18
Step 1
1 of 3
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi-annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{Ken started investing at age 20 with $ 100 per month}} hfill \
{text{at 5}}{text{.4% /a compounded monthly}}{text{. }} hfill \
{text{We want its value at age 55}} hfill \
R = $ 100 hfill \
i = frac{{0.054}}{{12}} = 0.0045 hfill \
m = left( {55 – 20} right) times 12 = 420 hfill \
hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} = 100 cdot frac{{{{left( {1 + 0.0045} right)}^{420}} – 1}}{{0.0045}} hfill \
FV = 124,252.52 hfill \
hfill \
{text{His twin brother started at age 37 }} hfill \
{text{at 7}}{text{.2% /a compounded monthly}} hfill \
{text{We want to know how much should he invest per month}} hfill \
{text{so that its value at age 55 is the same with Ken’s}} hfill \
hfill \
i = frac{{0.072}}{{12}} = 0.006 hfill \
m = left( {55 – 37} right) times 12 = 216 hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} Rightarrow R = frac{{FV cdot i}}{{{{left( {1 + i} right)}^m} – 1}} hfill \
R = frac{{124,252.52left( {0.006} right)}}{{{{left( {1 + 0.006} right)}^{ – 216}} – 1}} = $ 282.34 hfill \
hfill \
{text{difference = 282}}{text{.34}} – 100 = boxed{{mathbf{$ 182}}{mathbf{.34}}} hfill \
hfill \
{text{Therefore, Adam must invest $ 182}}{text{.34 more than Ken}}{text{.}} hfill \
end{gathered} ]
Result
3 of 3
$$
$182.34
$$
Exercise 19
Step 1
1 of 5
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 5
[begin{gathered}
{text{In this case,}} hfill \
PV = $ 100,000 hfill \
r = 4.2% /a hfill \
i = frac{{0.042}}{{12}} = 0.0035 hfill \
m = 12 times t hfill \
{text{compounded monthly}} hfill \
hfill \
{text{We must compare the amount of the time}} hfill \
{text{that she can pay off the loan if }} hfill \
R = 1000{text{ and }}R = 1500 hfill \
hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} Rightarrow frac{{PV}}{R} = frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
{text{If }}R = 1000,, Rightarrow ,,frac{{100,000}}{{1000}} = frac{{1 – {{left( {1 + 0.0035} right)}^{ – 12t}}}}{{0.0035}} hfill \
{text{If }}R = 1500,, Rightarrow ,,frac{{100,000}}{{1500}} = frac{{1 – {{left( {1 + 0.0035} right)}^{ – 12t}}}}{{0.0035}} hfill \
end{gathered} ]
Step 3
3 of 5
Exercise scan
Step 4
4 of 5
Therefore, if Jenny pays $1000 per month, it would take 10.275 years (10 years and 4 months) while if she pays $1500, it would just take 6.337 years (6 years and 4 months).

Thus paying the maximum amount would take 4 years earlier.

Result
5 of 5
4 years
Exercise 20
Step 1
1 of 3
[begin{gathered}
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
hfill \
{text{where}} hfill \
hfill \
PV = {text{ present value}} hfill \
R = {text{ regular payments}} hfill \
i = {text{ interest rate per compounding period}} hfill \
m = {text{ number of compounding periods}} hfill \
hfill \
{text{Normally, annual interest rate }}r{text{ and compounding mode is given}} hfill \
{text{annually }} Rightarrow {text{ }}i = r hfill \
{text{semi – annually }} Rightarrow {text{ }}i = frac{r}{2} hfill \
{text{quarterly }} Rightarrow {text{ }}i = frac{r}{4} hfill \
{text{monthly }} Rightarrow {text{ }}i = frac{r}{{12}} hfill \
hfill \
{text{The number of compounding period }}m,,{text{is the product of}} hfill \
{text{number of compounding per year }}n{text{ and the number of years}};t. hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{In this case,}} hfill \
R = $ 17.85 hfill \
r = 13% /a hfill \
{text{compounded weekly}} hfill \
i = frac{{0.13}}{{52}} = 0.0025 hfill \
m = {text{ 2}}{text{.5}} times 52 = 130 hfill \
{text{compounded weekly}} hfill \
hfill \
{text{We must find }}PV hfill \
hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
PV = 17.85 cdot frac{{1 – {{left( {1 + 0.0025} right)}^{ – 130}}}}{{0.0025}} = boxed{{mathbf{$ 1979}}{mathbf{.06}}} hfill \
hfill \
{text{Therefore, the selling price of the guitar is $ 1,979}}{text{.06}}{text{.}} hfill \
end{gathered} ]
Result
3 of 3
$$
$1,979.06
$$
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